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svp [43]
3 years ago
12

A ball thrown horizontall at 22.2 m/s from the roof of a building lands 36.0m from the base of the building. How tall is the bui

lding?
Physics
1 answer:
vekshin13 years ago
8 0
Velocity of the ball (v) = 22.2 m/s
Distance from the base of the building where the ball falls (d) = 36.0m
We know that, Velocity (v) = distance (d) / time (t)
From the above equation,
Time (t) = distance (d) / velocity (v)
Substituting the values for distance and velocity in the above equation,
Time = 36.0/22.2 = 1.62 seconds

When the ball is falling down, it accelerates due to gravity
This acceleration (a) = 9.8 m/s²

Now we need to find the height of the building.
The equation of motion for an object with constant acceleration is,
s = ut + 1/2 at²
Where,
s = vertical displacement, which is height of the building in our case
u = initial velocity of the ball = 0

Substituting all the above values in the equation of motion,
s = (0 × 1.62) + (1/2 × 9.8 × 1.62²)
s = 0 + (25.72/2) = 12.86 m = 12.9 m which is the height of the building

Therefore the building is 12.9m tall

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
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Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

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